/************************************************************************************

    Solving Diophantine equations in form of a * x + b * y = c

    Uses extended Euclid algorithm 
    (which finds such x, y that a * x + b * y = gcd(a, b))

    Based on problem 4188 from informatics.mccme.ru
    http://informatics.mccme.ru/moodle/mod/statements/view3.php?chapterid=4188

************************************************************************************/

#include <iostream>
#include <fstream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cassert>
#include <utility>
#include <iomanip>

using namespace std;

long long gcd(long long a, long long b) {
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// Finds such x and y, that a * x + b * y = gcd(a, b)
long long extended_gcd(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
        x = 1; y = 0; 
        return a;
    }
    long long x1, y1;
    long long g = extended_gcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - (a / b) * y1;
    return g;
}

// Solves equation a * x + b * y = c, writes answer to x and y
void solveDiophantine(long long a, long long b, long long c, long long &x, long long &y) {
    long long g = extended_gcd(a, b, x, y);

    if (c % g != 0) {
        puts("Impossible");
        exit(0);
    }

    c /= g;

    x = x * c; y = y * c;
}

long long a, b, c;
long long x, y;
long long g;

int main() {
    //assert(freopen("input.txt","r",stdin));
    //assert(freopen("output.txt","w",stdout));

    cin >> a >> b >> c;

    // Find any solution
    solveDiophantine(a, b, c, x, y);

    // In this problem we search for solution with minimum x >= 0
    // a * x + b * y = gcd(a, b)
    // now for any integer k: a * (x + k * b / g) + b * (y - k * a / g) = gcd(a, b)

    g = gcd(a, b);

    long long add = b / g;
    long long num = 0;
    if (add < 0) 
        num = (long long) floor(1.0 * -x / add);
    else
        num = (long long) ceil(1.0 * -x / add);

    x = x + b / g * num;
    y = y - a / g * num;

    cout << x << " " << y;

    return 0;
}
